Question

The polar of $(-3,5)$ w.r.t $x^2+y^2-2x-4y+3=0$ cuts the circle, its mid point is

• A. $(\frac{-17}{25},\frac{56}{25})$
• B. $(\frac{-17}{25},\frac{-56}{25})$
• C. $(\frac{17}{25},\frac{56}{25})$
• D. $(\frac{17}{25},\frac{-56}{25})$

Answer 1

The correct option is C $(\frac{17}{25},\frac{56}{25})$

Polar of $(-3,5)$ wrt $x^2+y^2-2x-4y+3=0$ is chord of contact from $(-3,5)$ to a given circle.
$\therefore x(-3)+5y-(x-3)-2(y+5)+3=0$
$-4x+3y-4=0$
$4x-3y+4=0$ -----(1)
mid-point of polar of $(-3,5)$ is foot perpendicular to chord $4x-3y+4=0$ from centre of circle
$\frac{x-1}{4}=\frac{y-2}{-3}=-\frac{\left(2\right)}{4^2+3^2}$
$x=1-\frac{8}{25}=\frac{17}{25}$
and $y=2+\frac{6}{25}=\frac{65}{25}$
$\Rightarrow (\frac{17}{25},\frac{56}{25})$ is a mid-point of polar of $(-3,5)$